Question: Line segment $\overline{AB}$ is extended past $B$ to $P$ such that $AP:PB = 10:3.$  Then
\[\overrightarrow{P} = t \overrightarrow{A} + u \overrightarrow{B}\]for some constants $t$ and $u.$  Enter the ordered pair $(t,u).$

[asy]
unitsize(1 cm);

pair A, B, P;

A = (0,0);
B = (5,1);
P = interp(A,B,10/7);

draw(A--P);

dot("$A$", A, S);
dot("$B$", B, S);
dot("$P$", P, S);
[/asy]
Since $AP:PB = 10:3,$ we can write
\[\frac{\overrightarrow{P} - \overrightarrow{A}}{10} = \frac{\overrightarrow{P} - \overrightarrow{B}}{7}.\]Isolating $\overrightarrow{P},$ we find
\[\overrightarrow{P} = -\frac{3}{7} \overrightarrow{A} + \frac{10}{7} \overrightarrow{B}.\]Thus, $(t,u) = \boxed{\left( -\frac{3}{7}, \frac{10}{7} \right)}.$